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CIE 9709 2026 June Paper 11 Q5

A Level / CIE / P1

CIE 9709 2026 June Paper 11 Paper · Question 5

题目

Problem

The equation of a curve is

y=−12x+x3−3x−2.y=-\frac{1}{2}x+\sqrt{x^3-3x-2}.

Find an equation of the normal to the curve at the point (6,11)(6,11). Give your answer in the form ax+by+c=0ax+by+c=0, where aa, bb and cc are integers to be found.

(5)
题目中文翻译

一条曲线的方程为

y=−12x+x3−3x−2.y=-\frac{1}{2}x+\sqrt{x^3-3x-2}.

求该曲线在点 (6,11)(6,11) 处的法线方程。答案写成 ax+by+c=0ax+by+c=0 的形式,其中 aa、bb 和 cc 是待求的整数。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

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对根式用链式法则。代入切点求切线斜率,再取负倒数求法线。

答题过程

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The derivative is

y′=−12+3x2−32x3−3x−2.y'=-\frac12+\frac{3x^2-3}{2\sqrt{x^3-3x-2}}.

At x=6x=6, the square root is 1414, so y′=−12+10528=134y'=-\frac12+\frac{105}{28}=\frac{13}4. The normal slope is −4/13-4/13. Its equation is y−11=−413(x−6)y-11=-\frac4{13}(x-6), or

4x+13y−167=0.4x+13y-167=0.